PSYCH/625 Version 2 Time to Practice � Week Four University of Phoenix Material

Name: PSYCH/625 Version 2 Time to Practice � Week Four University of Phoenix Material
Brand: University of Phoenix
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Author: Benjamin Clark

A set of practice problems designed to reinforce psychological concepts from Week Four of PSYCH/625.

Benjamin Clark

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Time to Practice – Week Four
PSYCH/625 Version 2
1
Copyright © 2014 by University of Phoenix. All rights reserved.
PSYCH/625 Version 2 Time to Practice – Week Four
University of Phoenix Material
Time to Practice – Week Four
Complete Parts A, B, and C below.
Part A
Some questions in Part A require that you access data from Statistics for People Who (Think They) Hate
Statistics. This data is available on the student website under the Student Text Resources link.
1. Using the data in the file named Ch. 11 Data Set 2, test the research hypothesis at the .05 level of
significance that boys raise their hands in class more often than girls. Do this practice problem by
hand using a calculator. What is your conclusion regarding the research hypothesis? Remember to
first decide whether this is a one- or two-tailed test.
This is a one-tailed test because you want to determine if the mean of one group (boys) – μ1 -
is higher than the mean of the second group (girls) – μ2.
Null hypothesis: μ1 = μ2
Alternative hypothesis: μ1 > μ2
Boys X - Mean (X-Mean)2
9 1.071429 1.147959
8 0.071429 0.005102
4 -3.92857 15.43367
9 1.071429 1.147959
3 -4.92857 24.29082
8 0.071429 0.005102
10 2.071429 4.290816
8 0.071429 0.005102
9 1.071429 1.147959
8 0.071429 0.005102
10 2.071429 4.290816
7 -0.92857 0.862245
6 -1.92857 3.719388
12 4.071429 16.57653
Sum = 111 Sum = 72.93
Mean = 7.93
S2 = 72.93 / 13 = 5.61
Girls X - Mean (X-Mean)2
3 -2.3125 5.347656
5 -0.3125 0.097656
1 -4.3125 18.59766
2 -3.3125 10.97266
6 0.6875 0.472656
4 -1.3125 1.722656

Time to Practice – Week Four
PSYCH/625 Version 2
2
Copyright © 2014 by University of Phoenix. All rights reserved.
3 -2.3125 5.347656
6 0.6875 0.472656
7 1.6875 2.847656
9 3.6875 13.59766
7 1.6875 2.847656
3 -2.3125 5.347656
7 1.6875 2.847656
6 0.6875 0.472656
8 2.6875 7.222656
8 2.6875 7.222656
Sum = 85 Sum = 85.44
Mean = 5.31
S2 = 85.44 / 15 = 5.69
Degrees of Freedom Total = 16 + 14 – 2 = 28
S2 pooled = 5.61 (13 / 28) + 5.69 (15 / 28) = 5.658
Variance for Boys: S2 = 5.658 / n = 5.658 / 14 = 0.404
Variance for Girls: S2 = 5.658 / n = 5.658 / 16 = 0.354
S2 pop = 0.354 + 0.404 = 0.756
S pop = sqrt 0.756 = 0.869
T = (7.93 – 5.31) / 0.869 = 3.01
Looking at the critical values tables, the t-value at 28 degrees of freedom is
1.701 at p = 0.05 for a two-tailed test. The calculated t-value is larger so the
test is significant.
An independent samples t-test showed that there was a significant difference
between the mean number of times boys raised their hands in class versus
girls, t(28) = 3.01, 0.005 > p > 0.0025 for a two-tailed test. The p-value would be
0.0025 > p > 0.00125 for a one-tailed test. The boys would raise their hands
more in class (M = 7.93, SD = 2.37) than girls (M = 5.31, SD = 2.39). We should
reject the null.
2. Practice the following problems by hand just to see if you can get the numbers right. Using the
following information, calculate the t test statistic.
a. 𝑋1 = 62 𝑋2 = 60 𝑛1 = 10 𝑛2 = 10 𝑠1
2 = 2.45 𝑠2
2 = 3.16
Find the pooled S2 = 2.45 (9 / 18) + 3.16 (9 / 18) = 2.45
Variance for 1 = 2.45 / 10 = 0.245
Variance for 2 = 0.245
Pop Variance = 0.245 + 0.245 = 0.49

Time to Practice – Week Four
PSYCH/625 Version 2
3
Copyright © 2014 by University of Phoenix. All rights reserved.
Sqrt 0.49 = 0.7
T = (62 – 60) / 0.7 = 2.86
b. 𝑋1 = 158 𝑋2 = 157.4 𝑛1 = 22 𝑛2 = 26 𝑠1
2 = 2.06 𝑠2
2 = 2.59
Find the pooled S2 = 2.06 (21 / 46) + 2.59 (25 / 48) = 2.35
Variance for 1 = 2.35 / 22 = 0.107
Variance for 2 = 2.35 / 26 = 0.0903
Pop Variance = 0.107 + 0.0903 = 0.197
Sqrt 0.197 = 0.444
T = (158 – 157.4) / 0.444 = 1.35
c. 𝑋1 = 200 𝑋2 = 198 𝑛1 = 17 𝑛2 = 17 𝑠1
2 = 2.45 𝑠2
2 = 2.35
Find the pooled S2 = 2.45 (16 / 32) + 2.35 (16 / 32) = 2.40
Variance for 1 = 2.45 / 17 = 0.144
Variance for 2 = 2.35 / 17 = 0.138
Pop Variance = 0.144 + 0.138 = 0.282
Sqrt 0.282 = 0.531
T = (200 - 198) / 0.531 = 3.77
3. Using the results you got from Question 2 and a level of significance at .05, what are the two-tailed
critical values associated with each? Would the null hypothesis be rejected?
a) Critical value at df = 18 and alpha = 0.05, t = 1.734. The null hypothesis
would be rejected because the calculated t = 2.86 is greater than critical t =
1.734
b) Critical value at approximate df = 40 and alpha = 0.05, t = 1.684. The null
hypothesis would not be rejected because the calculated t = 1.35 is less
than critical t = 1.684
c) Critical value at approximate df = 30 and alpha = 0.05, t = 1.697. The null
hypothesis would be rejected because the calculated t = 3.77 is greater
than critical t = 1.697
4. Using the data in the file named Ch. 11 Data Set 3, test the null hypothesis that urban and rural
residents both have the same attitude toward gun control. Use IBM® SPSS® software to complete the
analysis for this problem.
Group Statistics
Group N Mean Std. Deviation Std. Error Mean
Attitude Urban 16 6.5112 1.77221 .44305
Rural 14 5.3979 3.31442 .88582

Time to Practice – Week Four
PSYCH/625 Version 2
4
Copyright © 2014 by University of Phoenix. All rights reserved.
Independent Samples Test
Levene's Test for
Equality of
Variances t-test for Equality of Means
F Sig. t df
Sig. (2-
tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower Upper
Attitude Equal variances
assumed
4.463 .044 1.168 28 .253 1.11339 .95311 -.83897 3.06576
Equal variances
not assumed
1.124 19.273 .275 1.11339 .99044 -.95763 3.18442
An independent samples t-test was performed on the sample. The mean attitude for urban
residents (M = 6.51, SD = 1.77) was slightly higher than the mean attitude for rural residents (M
= 5.40, SD = 3.31). The Levene’s Test was significant, F = 4.463, p = 0.044, which suggests that
the variances of the two groups are not equal. The t-test was not significant, t (19) = 1.124, p =
0.274. There is no difference in attitude on gun control among urban and rural residents. We
would fail to reject the null.
5. In the following examples, indicate whether you would perform a t test of independent means or
dependent means.
a. Two groups were exposed to different treatment levels for ankle sprains. Which treatment was
most effective? A t-test of independent means would be used because the two groups were
mutually exclusive and had different subjects in each group.
b. A researcher in nursing wanted to know if the recovery of patients was quicker when some
received additional in-home care whereas when others received the standard amount. A t-test of
independent means was used because the groups were independent of each other.
c. A group of adolescent boys was offered interpersonal skills counseling and then tested in
September and May to see if there was any impact on family harmony. A t-test of dependent
means was used because the same group of boys was tested again at two different time
points.
d. One group of adult men was given instructions in reducing their high blood pressure whereas
another was not given any instructions. A t-test of independent means was used because two
different groups of men are being tested.
e. One group of men was provided access to an exercise program and tested two times over a 6-
month period for heart health. A t-test of dependent means was used because the same
group of men was tested twice.
6. The data set for this problem can be found through the Sage Materials in the Student Textbook

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Subject

Psychology

PSYCH/625 Version 2 Time to Practice � Week Four University of Phoenix Material

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